Positive solutions of the three-point boundary value problem for fractional-order differential equations with an advanced argument

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1 RESEARCH Open Access Positive solutions of the three-point boundary value problem for fractional-order differential equations with an advanced argument Guotao Wang *, SK Ntouyas 2 and Lihong Zhang * Correspondence: wgt252@63. com School of Mathematics and Computer Science, Shanxi Normal University, Linfen, Shanxi 44, P. R. China Full list of author information is available at the end of the article Abstract In this article, we consider the existence of at least one positive solution to the three-point boundary value problem for nonlinear fractional-order differential equation with an advanced argument { C D α u(t)+a(t)f (u(θ(t))), < t <, u() u (), u() u(), where 2 <a 3, <h <, <<, C D a is the Caputo fractional derivative. Using the well-known Guo-Krasnoselskii fixed point theorem, sufficient conditions for the existence of at least one positive solution are established. MSC (2): 34A8; 34B8; 34K37. Keywords: Positive solution, Three-point boundary value problem, Fractional differential equations, Guo-Krasnoselskii fixed point theorem, Cone Introduction The study of three-point BVPs for nonlinear integer-order ordinary differential equations was initiated by Gupta []. Many authors since then considered the existence and multiplicity of solutions (or positive solutions) of three-point BVPs for nonlinear integer-order ordinary differential equations.toidentifyafew,wereferthereaderto [2-3] and the references therein. Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of complex medium, polymer rheology, etc. [4-7]. In fact, fractional-order models have proved to be more accurate than integerorder models, i.e., there are more degrees of freedom in the fractional-order models. In consequence, the subject of fractional differential equations is gaining much importance and attention. For details, see [8-36] and the references therein. Differential equations with deviated arguments are found to be important mathematical tools for the better understanding of several real world problems in physics, mechanics, engineering, economics, etc. [37,38]. In fact, the theory of integer order differential equations with deviated arguments has found its extensive applications in realistic mathematical modelling of a wide variety of practical situations and has emerged 2 Wang et al; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

2 Page 2 of as an important area of investigation. For the general theory and applications of integer order differential equations with deviated arguments, we refer the reader to the references [39-45]. As far as we know, fractional order differential equations with deviated arguments have not been much studied and many aspects of these equations are yet to be explored. For some recent work on equations of fractional order with deviated arguments, see [46-48] and the references therein. In this article, we consider the following three-point BVPs for nonlinear fractional-order differential equation with an advanced argument { C D α u(t)+a(t)f (u(θ(t))), < t <, u() u (:) (), u() u(), where 2 <a 3, <h <, <<,C D a is the Caputo fractional derivative and f : [, ) [, ) is a continuous function. By a positive solution of (.), one means a function u(t) that is positive on <t< and satisfies (.). Our purpose here is to give the existence of at least one positive solution to problem (.), assuming that (H ): a Î C ([, ], [, )) and a does not vanish identically on any subinterval. (H 2 ): The advanced argument θ Î C((, ), (, )) and t θ(t), t Î (, ). Let E C[, ] be the Banach space endowed with the sup-norm. Set f (u) f lim u + u, f f (u) lim u u. The main results of this paper are as follows. Theorem. Assume that (H ) and (H 2 ) hold. If f and f, then problem (.) has at least one positive solution. Theorem.2 Assume that (H ) and (H 2 ) hold. If f and f, then problem (.) has at least one positive solution. Remark. It is worth mentioning that the conditions of our theorems are easily to verify, so they are applicable to a variety of problems, see Examples 4. and 4.2. The proof of our main results is based upon the following well-known Guo-Krasnoselskii fixed point theorem: Theorem.3 [49] LetEbeaBanachspace,andletP Ebeacone.Assumethat Ω, Ω 2 are open subsets of E with Î Ω, 2, and let T : P ( 2 \ ) Pbe a completely continuous operator such that (i) Tu u, u Î P Ω, and Tu u, u Î P Ω 2 ; or (ii) Tu u, u Î P Ω, and Tu u, u Î P Ω 2. Then T has a fixed point in P ( 2 \ ). 2 Preliminaries For the reader s convenience, we present some necessary definitions from fractional calculus theory and Lemmas. Definition 2. For a function f : [, ) R, the Caputo derivative of fractional order a is defined as

3 Page 3 of C D α t f (t) (t s) n α f (n) (s)ds, n <α<n, n [α]+, Ɣ(n α) where [a] denotes the integer part of real number a. Definition 2.2 The Riemann-Liouville fractional integral of order a is defined as I α f (t) t (t s) α f (s)ds, α>, provided the integral exists. Definition 2.3 The Riemann-Liouville fractional derivative of order a for a function f (t) is defined by D α f (t) Ɣ(n α) ( d t dt )n (t s) n α f (s)ds, n [α]+, provided the right-hand side is pointwise defined on (, ). Lemma 2. [5] Let a >, then fractional differential equation C D α u(t) has solution u(t) C + C t + C 2 t C n t n, C i R, i,,2,..., n, where n is the smallest integer greater than or equal to a. Lemma 2.2 [5] Let a >, then I αc D α u(t) u(t) + C + C t + C 2 t C n t n for some C i Î R, i,,2,...,n-, where N is the smallest integer greater than or equal to a. Lemma 2.3 Let 2 < a 3, bh. Assume y(t) Î C[, ], then the following problem C D α u(t) + y(t), < t <, (2:) u() u (), u() u(), (2:2) has a unique solution t u(t) (t s) α y(s)ds+ t( s) α y(s)ds t( s) α y(s) ds. Proof. We may apply Lemma 2.2 to reduce Equation (2.) to an equivalent integral equation t u(t) I α y(t) b b 2 t b 3 t 2 for some b, b 2, b 3 Î R. (t s) α y(s)ds b b 2 t b 3 t 2, (2:3)

4 Page 4 of In view of the relation C D a I a u(t) u(t) andi a I b u(t) I a+b u(t) fora, b >, we can get that t u (t) t u (t) (t s) α 2 Ɣ(α ) y(s)ds b 2 2b 3 t. (t s) α 3 Ɣ(α 2) y(s)ds 2b 3. By u() u (), it follows b b 3. Then by the condition bu(h) u(), we have b 2 Combine with (2.3), we get t u(t) This complete the proof. ( s) α y(s)ds (t s) α y(s)ds+ ( s) α y(s) ds. t( s) α y(s)ds t( s) α y(s) ds. Lemma 2.4 Let 2 < a 3, <<. Assume y Î C([, ], [, )), then the unique solution u of (2.) and (2.2) satisfies u(t), t Î [, ]. Proof. By Lemma 2.3, we know that u (t) t the graph of u(t) is concave down on (, ). In addition, u() ( s) α y(s)ds + ( s) α y(s)ds ( ) ( ) ( s) α y(s)ds ( s) α y(s)ds (t s) α 3 y(s)ds. Itmeansthat Ɣ(α 2) ( s) α y(s)ds ( s) α y(s)ds [ ( s) α ( s) α ] y(s)ds [ ( s) α ( s) α ] y(s)ds Combine with u(), it follows u(t), t Î [, ]. ( s) α y(s)ds Lemma 2.5 Let 2 < a 3, <<. Assume y Î C([, ], [, )), then the unique solution u of (2.) and (2.2) satisfies

5 Page 5 of inf u(t) γ u, t [,] ( ) where γ min{,, }. Proof. Notethatu (t), by applying the concavity of u, theproofiseasy.sowe omit it. 3 Proofs of main theorems Define the operator T : C[, ] C[, ] as follows, t Tu(t) (t s) α t( s) α. t( s) α (3:) Then the problem (.) has a solution if and only if the operator T has a fixed point. Define the cone P {u u C[, ], u, inf u(θ(t)) γ u }, where t [,] ( ) γ min{,, }. ProofofTheorem.. The operator T is completely continuous. Obviously, T is continuous. Let Ω C[, ] be bounded, then there exists a constant K>suchthat a(t) f (u (θ(t)) K, u Î Ω. Thus, we have Tu (t) ( s) α ( s) α ds K K ( )Ɣ(α +), K which implies Tu ( )Ɣ(α +). On the other hand, we have t (Tu) (t) + K (t s) α 2 Ɣ(α ) ( s) α 2 Ɣ(α ) ds + ( s) α K K + ( + )K : M. ( )Ɣ(α +) ( s) α ds + ( s) α K ( s) α ds

6 Page 6 of Hence, for each u Î Ω, let t, t 2 Î [, ], t <t 2, we have (Tu)(t 2 ) (Tu)(t ) t 2 t (Tu) (s) ds M(t2 t ). So, T is equicontinuous. The Arzela-Ascoli Theorem implies that T : C[, ] C[, ] is completely continuous. Since t θ (t), t Î (, ), then inf u(θ(t)) inf u(t) γ u. (3:2) t [,] t [,] Thus, Lemmas 2.5 and 3.2 show that TP P. Then,T : P P is completely continuous. In view of f, there exists a constant r > such that f(u) δ u for <u<r, where δ > satisfies δ γ ( ) ( s) α a(s)ds. (3:3) Take u Î P, such that u r. Then, we have Tu Tu() ( s) α ( s) α ( s) α ( s) α ( s) α ( s) α ( s) α ( s) α + ( s) α ( s) α ( s) α + ( s) α ( s) α a(s)δ u(θ(s))ds ( s) α a(s)δ γ u ds δ γ ( s) α a(s)ds u u. ( ) (3:4) Let Ω r {u Î C [] u <r }. Thus, (3.4) shows Tu u, u P ρ.

7 Page 7 of Next, in view of f, there exists a constant R>r such that f(u) δ 2 u for u R, where δ 2 > satisfies δ 2 ( ) ( s) α a(s)ds. (3:5) We consider the following two cases. Case one. f is bounded, which implies that there exists a constant r > such that f (u) r for u Î [, ). Now, we may choose u Î P such that u r 2,wherer 2 max {μ, R}. Then we have t Tu (t) (t s) α t( s) α ( s) α r ( s) α a(s)ds ( ) μ ρ 2 u. t( s) α Case two. f is unbounded, which implies then there exists a constant ρ 2 > R γ > R such that f(u) f(r 2 )for<u r 2 (note that f Î C([, ), [, )). Let u Î P such that u r 2, we have Tu (t) δ 2 ( ) u. ( s) α ( s) α a(s)f (ρ 2 )ds ( s) α a(s)δ 2 ρ 2 ds ( s) α a(s)ds u Hence, in either case, we may always let Ω r2 {u Î C[, ] u <r 2 } such that Tu u for u P ρ2. Thus, by the first part of Guo-Krasnoselskii fixed point theorem, we can conclude that (.) has at least one positive solution. Proof of Theorem.2. Now, in view of f, there exists a constant r > such that f (u) τ u for <u<r, where τ > satisfies

8 Page 8 of τ ( ) ( s) α a(s)ds. (3:6) Take u Î P, such that u r. Then, we have ( s) α Tu (t) ( s) α a(s)τ u(θ(s))ds τ ( s) α a(s)ds u ( ) u. (3:7) Let Ω {u Î C[, ] u <r }. Thus, (3.7) shows Tu u, u Î P Ω. Next, in view of f, there exists a constant r 2 >r such that f(u) τ 2 u for u r 2, where τ 2 > satisfies τ 2 γ ( ) ( s) α a(s)ds. (3:8) Let Ω 2 {u Î C[, ] u <r 2 },where ρ 2 > r 2 γ > r 2,then,u Î P and u r 2 implies and so inf u(θ(t)) γ u > r 2, t [,] Tu Tu() τ 2 γ ( ) ( s) α ( s) α a(s)τ 2 u(θ(s))ds ( s) α a(s)τ 2 γ u ds ( s) α a(s)ds u u. This shows that Tu u for u Î P Ω 2. Therefore, by the second part of Guo-Krasnoselskii fixed point theorem, we can conclude that (.) has at least one positive solution u P ( 2 \ ).

9 Page 9 of 4 Examples Example 4. Consider the fractional differential equation { C D α u(t)+e t f (u(θ(t))), < t <, u() u (), u() u(), (4:) where 2<a 3, <h <, <<, θ(t) tv,<v< and sin u u 2, u π 2, f (u) 4 2u +cosu, u > π 5 2. π 2 Note that conditions (H )and(h 2 ) of Theorem. hold. Through a simple calculation we can get f and f. Thus, by Theorem., we can get that the problem (4.) has at least one positive solution. Example 4.2 Consider the fractional differential equation { C D α u(t)+a(t)f (u(θ(t))), < t <, u() u (), u() u(), (4:2) where 2<a 3, <h <, <<, θ(t) t, a(t) e tan t and 3 f (u) u2 ln(+u) + u 3+sin u. Obviously, it is not difficult to verify conditions (H )and(h 2 )oftheorem.2hold. Through a simple calculation we can get f and f. Thus, by Theorem.2, we can get that the problem (4.2) has at least one positive solution. Remark 4. In the above two examples, a, b, h could be any constants which satisfy 2<a 3, <h <, <<. For example, we can take a 2.5, h.5, b.5. Author details School of Mathematics and Computer Science, Shanxi Normal University, Linfen, Shanxi 44, P. R. China 2 Department of Mathematics, University of Ioannina, Ioannina 45, Greece Authors contributions GW completed the main part of this paper, SKN and LZ corrected the main theorems and gave two examples. All authors read and approved the final manuscript. Competing interests The authors declare that they have no competing interests. Received: 2 December 2 Accepted: 7 May 2 Published: 7 May 2 References. Gupta CP: Solvability of a three-point nonlinear boundary value problem for a second order ordinary differential equation. J Math Anal Appl 992, 68: Jankowski T: Positive solutions for three-point one-dimensional p-laplacian boundary value problems with advanced arguments. Appl Math Comput 29, 25: Jankowski T: Solvability of three point boundary value problems for second order differential equations with deviating arguments. J Math Anal Appl 25, 32: Ma R: Multiplicity of positive solutions for second-order three-point boundary value problems. Comput Math Appl 2, 4:93-24.

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